Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:

import java.util.Scanner;

/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author aakas
 */
public class Main{
    static int n;
    static double[] p;
    static boolean v[][];
    static double dp[][];
    public static void main(String args[])
    {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        p = new double[n];
        v = new boolean[n+1][n+1];
        dp = new double[n+1][n+1];
        for(int i=0;i<n;i++)
        {
            p[i] = sc.nextDouble();
        }
        double sum = 0;
        for(int i=0;i<Math.ceil((double)n/2.0);i++)
        {
            sum += s(n-i,i);
        }
        System.out.println(sum);
    }
    public static double s(int i,int j)
    {
        if(i==0 && j==1)
            return 1-p[i+j-1];
        if(i==1 && j==0)
            return p[i+j-1];
        if(i==0 && j==0)
            return 0;
        if(i<0 || j<0)
            return 0;
        if(v[i][j])
            return dp[i][j];
        v[i][j] = true;
        return dp[i][j] = p[i+j-1]*s(i-1,j) + (1-p[i+j-1])*(s(i,j-1)); 
    }
}